# Computing $\lim_{n\rightarrow \infty} \frac{1+\sqrt{2}+\sqrt[3]{3}+\cdots+\sqrt[n]{n}}{n}$.

I have a problem with the calculation of the following limit. $$\lim_{n\rightarrow \infty} \frac{1+\sqrt{2}+\sqrt[3]{3}+\cdots+\sqrt[n]{n}}{n}$$ I do not know where to start! Thank you very much

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Do you know the limit of $n^{1/n}$? Also $n=\sum_{i=1}^n 1$. –  Johan Nov 19 '12 at 13:54
Use $\sqrt[n]{n}\to1$, see e.g. here, and Cesaro mean. –  Martin Sleziak Nov 19 '12 at 14:12
As $n$ grows large, what you do when you co from case $n$ to case $n+1$ is adding $1$ to the denominator and adding ever so slightly more than $1$ to the numerator. Intuition dictates it tends towards $1$, and once you have a convergence candidate, you're halfway there. –  Arthur Nov 19 '12 at 14:17

There are at least two possibilities. In all of them you use that $\lim_{n\to\infty}=\sqrt[n]{n}=1$.

The first one uses the following result: if $a_n$ is a convergent sequence, then $$\lim_{n\to\infty}\frac{a_1+a_2+\dots+a_n}{n}=\lim_{n\to\infty}a_n.$$

The second is to use Stolz's criterion.

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I remember that there's some theorem saying that, given 2 positive sequences $(a_n); (b_n)$ if $\lim \frac{a_n}{b_n} = \alpha$, then $\lim \frac{\mathop\sum_{i = 1}^n a_i}{\mathop\sum_{i = 1}^n b_i} = \alpha$. Is this correct, and what's the name of this theorem? –  user49685 Nov 19 '12 at 14:27
@user49685 It is (a version of) Stolz-Cesaro theorem, see e.g. this answer. –  Martin Sleziak Nov 19 '12 at 14:34

Hint: Note that $$\sqrt[n]{n} \rightarrow 1.$$

You'll need to know and use that fact.

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@DonAntonio I second that. –  Dan Shved Nov 19 '12 at 14:08

Hint: It's easy. 1000^(1/1000) ~ 1. Use the algebra of limits: lim n-> infinity (a_n/b_n) = (lim n->infinity a_n)/(lim n->infinity b_n). The numerator tends to n (plus some constant) and the denominator tends to n. So the limit is 1.

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That is incorrect. $\lim_{n \rightarrow \infty} \frac{a_n}{b_n} = \frac{\lim_{n \rightarrow \infty}a_n}{\lim_{n \rightarrow \infty}b_n}$ if both limits are finite and denominator limit is not 0. Also What does "numerator tend to n" mean when you take limit to infinity? –  Gautam Shenoy Nov 19 '12 at 14:06
I mean you get S_n = 1 + (close to 1) + (close to 1) + ... n times which is approximately n. If you want more rigour, I only need to mention that n^(1/n) tends to 1 as n tends to infinity. That way, given epsilon, there exists an n_0 such that n^(1/n) - 1 is less than epsilon for all n >= n_0. Or in other words, S_n gets relatively closer to n as n tends to infinity. When n is 1000 S_n is relatively close to n = 1000. When n is 1,000,000, S_n is relatively even closer to n = 1,000,000 than it was for n = 1000. This is all that matters here. –  Adam Rubinson Nov 19 '12 at 14:30
If (a_n) is a sequence with a_n-->1 as n-->infinity then lim(k-->infinity) [Sum(a_1 to a_k)] / k = 1. –  Adam Rubinson Nov 19 '12 at 14:37
@AdamRubinson Of course this result is true. The argument in your answer is rather poor, though. –  Cocopuffs Nov 19 '12 at 14:42
Sure. For stuff like this I think intuition is more important than rigour. But maybe that's personal taste –  Adam Rubinson Nov 19 '12 at 14:46